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a(n) will be the same as A219029(n) except when n is a member of A033949 or n = 1, i.e. n is not 2, 4, prime, power of a prime, twice a prime, or twice a prime power. In such cases, when n is a member of A033949, then a(n) = n-1.

Rows with indices n in A033948 (or with A046144(n)=0) are empty. Indices of nonempty rows are given by A033949.

This is A228179 without the trivial square roots {1, n-1}.

The number of terms in each nonempty row n is even: A060594(n)-2.

Terms are not of the form p^k, where p is a prime.

There are no terms of the form 2p+1, where p is a prime.

The sequence contains all Carmichael numbers except A264012.

If n is in the sequence, then n-1 is not squarefree. - Thomas Ordowski, Jan 02 2017

Theorem: the set of such numbers has natural density 0. Proof: Let y = y(n) = loglog n /logloglog n. Using part 1 of Lemma 2.1 in paper 199 on my home page (joint with Luca), applied to the residue class 1: But for a set of n of density 0, for each integer d < y, there is a prime p|n with p == 1 (mod d). In particular, lambda(n) is divisible by every integer d up to y. Suppose now that gcd(lambda(n),n-1) < gcd(phi(n),n-1). Then there is a prime power q^a such that q^a | phi(n), q^a | n-1, and q^a does not divide lambda(n). Then, but for a set of n of density 0, we would have q^a > y. Since q | lambda(n), we have a at least 2. So, n-1 is divisible by some q^a > y with a >= 2. The set of such n has density 0. QED. - Carl Pomerance, Jan 02 2017

Number of terms < 10^k: 0, 8, 112, 1258, 13069, 132262, 1324263, 13229372, 132009236, ..., . Robert G. Wilson v, Jan 04 2017

If p and q are distinct primes == 3 (mod 4), then p*q is in the sequence. - Thomas Ordowski, Mar 30 2017

Terms (1, 2, 4) followed by A033949, positive integers that do not have a primitive root.

Also numbers n for which A353768(n) and A353768(A267099(n)) are equal. Proof: if n is an odd prime power or twice such a number, then the odd prime factor in A267099(n) is in the opposite side of 4k+1 / 4k+3 divide of that of the odd prime factor of n, and subtracting one from it will give a number of the form 4k+0 in the other case, and 4k+2 in the other case, and either 4k != 4k+2 (mod 4) when the prime factor is unitary, or then 4k*(4k+1) != (4k+2)*(4k+3) (mod 4), when the odd prime has exponent > 1, so none of such n occur in this sequence. On the other hand, if n has more than two distinct odd prime factors, p and q, then (p-1)(q-1) == 0 (mod 4), or if n is a multiple of 4, then as phi(4) = 2 and phi(2^k) == 0 (mod 4) for k > 2, and with (p-1) giving at least one instance of factor 2, then both A267099(n) and n are guaranteed to be multiples of 4, regardless of whether p (and q) is (are) of the form 4k+1 or 4k+3.

Is a(n)<> 1 iff n in A033948, n>2? [R. J. Mathar, May 21 2009]

Same as A103131, except there -1 appears instead of n-1. By Gauss's generalization of Wilson's theorem, a(n)=-1 means n has a primitive root (n in A033948) and a(n)=1 means n has no primitive root (n in A033949). [T. D. Noe, May 21 2009]

The trivial solutions of the congruence x^2 == 1 (mod 3*prime(n+2)), n>=1, with the primes prime(n+2) = A000040(n+2) have positive representatives 1 and 3*prime(n+2)-1. There are all-together four incongruent solutions due to a general theorem (see, e.g., the Hardy-Wright reference, Theorem 122, p. 96, and also A060594) and the fact that the number of incongruent solutions of this congruence with odd prime modulus p is two, namely with positive representative p and p-1 (see, e.g., Hardy-Wright, Theorem 109, p. 85). a(n) is the smallest positive odd representative >1 which solves this congruence. The other nontrivial even representative solving this congruence is 3*prime(n+2) - a(n), i.e. 4, 8, 10, 14, 16, 20, ... See 2*A207336.

a(n) solves also the congruence x^2 == 1 (Modd A001748(n+2)), n>=1. For Modd n (not to be confused with mod n) see a comment on A203571. This follows from floor(a(n)^2/3*prime(n+2)) being even, in fact it is 8*A024699(n) (see a comment there), hence a(n)^2 (Modd 3*prime(n+2)) = a(n)^2 (mod 3*prime(n+2)) = 1. For those multiplicative groups Modd 3*p with p an odd prime which are cyclic (this is not possible in the mod case, see A033949), a(n) is the representative of the only other nontrivial solution of this congruence. The representative of the trivial solution is 1 (-1 belongs to the same Modd class). (The conjecture stated here earlier is wrong, that is, the multiplicative group Modd (91=7*13) is non-cyclic. It may still be true for 3*p. - Wolfdieter Lang, Mar 15 2012)

The multiplicative group of integers modulo n, (Z/n*Z)^x, also the cyclotomic group, the Galois group Gal(Q(zeta(n))/Q) with zeta(n) = exp(2*Pi*I/n), is cyclic for n from A033948 and non-cyclic for n from A033949. Each of these groups is the direct product of cyclic factors (one factor is included).

In the total factorization for n >= 3 only cyclic factors whose orders are prime powers appear, and because the direct product is associative, and for these abelian groups also commutative, one can order the factors with nonincreasing orders.

For n=1 and n=2 the group is C_1 = {1} (for n=1 one has 1 == 0 (mod 1)).

Cyclic groups may also have a factorization into more than one factor. E.g., C_6 = C_3 x C_2.

The number of factors in this total factorization is for a cyclic group C_m, for m >= 2, given by A001221(m). For m=1 this number is 1 (not A001221(1)).

For non-cyclic groups the number of factors in this total factorization is given by A281855(m) if n = A033949(m), m >= 1.

For the non-cyclic group case see also the W. Lang links under A281854.

Compare this sequence with A046072 where another factorization of these groups is used, the one with the least cyclic factors. E.g., A046072(7) = 1 for the group C_6, and a(7) = 2 here (see the example above).

Union of {0} and A033949.

Numbers n such that A046145(n)=0 except n=1.

Nonprime k such that the multiplicative group modulo k is cyclic. Nonprime terms of A033948 (omitting the initial term 1). - Joerg Arndt, Aug 07 2011

a(n) has a primitive root for any n. - Arkadiusz Wesolowski, Sep 06 2012 [See, e.g., the Niven et al. reference. - Wolfdieter Lang, Jan 18 2017]

There are exactly n - 1 - phi(phi(n)) non-primitive roots for n, less than n, if n is prime.

a(n) will be the same as A219027(n) except when n is a member of A033949 or n = 1, i.e., n is not 2, 4, prime, power of a prime, twice a prime, or twice a prime power.